An optical fibre having core of refractive index `sqrt3` and cladding of refractive index 1.5 is kept in air. The maximum angle of acceptance is

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `15^(@)`

Answer

To find the maximum angle of acceptance for an optical fiber with a core refractive index of \(\sqrt{3}\) and a cladding refractive index of \(1.5\), we can follow these steps:

### Step 1: Understand the Concept of Critical Angle
The maximum angle of acceptance in an optical fiber is related to the critical angle, which is the angle of incidence above which total internal reflection occurs. This angle can be calculated using Snell's law.

### Step 2: Apply Snell's Law
According to Snell's law:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
where:
- \(\mu_1\) is the refractive index of the core,
- \(\mu_2\) is the refractive index of the cladding,
- \(i\) is the angle of incidence (which will be the critical angle \(C\)),
- \(r\) is the angle of refraction (which is \(90^\circ\) at the critical angle).

### Step 3: Set Up the Equation
In our case:
- \(\mu_1 = \sqrt{3}\)
- \(\mu_2 = 1.5\)
- \(r = 90^\circ\) (so \(\sin(90^\circ) = 1\))

Thus, we can rewrite Snell's law as:
\[ \sqrt{3} \sin(C) = 1.5 \cdot 1 \]

### Step 4: Solve for \(\sin(C)\)
Rearranging the equation gives:
\[ \sin(C) = \frac{1.5}{\sqrt{3}} \]

### Step 5: Simplify the Expression
To simplify \(\frac{1.5}{\sqrt{3}}\):
\[ \sin(C) = \frac{1.5 \cdot \sqrt{3}}{3} = \frac{\sqrt{3}}{2} \]

### Step 6: Find the Critical Angle
Now, we need to find the angle \(C\) such that:
\[ \sin(C) = \frac{\sqrt{3}}{2} \]
This corresponds to:
\[ C = 60^\circ \]

### Step 7: Conclusion
The maximum angle of acceptance for the optical fiber is therefore:
\[ \text{Maximum angle of acceptance} = 60^\circ \]

### Final Answer
The maximum angle of acceptance is \(60^\circ\).

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